A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move
1 answer:
Answer:
(a) Region of higher potential
(b) ΔV=1.911V
Explanation:
For Part (a)
Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest
For Part (b)
Potential difference we can calculate bu definition of kinetic energy
ΔK=mv²/2= -qΔV
express in ΔV
ΔV= -mv²/2q
Substitute the given values
So
ΔV= -mv²/2q
ΔV=1.911V
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